Function Evaluation

[nycmath]

If y = sqrt{4 - x^2}, find

(a) y when x = y

(b) y when x = (1/x^10)

(c) y when x = - y

[Eigenvalue]

a) y=[tex]\sqrt{4-x²}[/tex]
y=x
x=[tex]\sqrt{4-x²}[/tex]
square both sides
x²=4-x²
2x²=4
x²=2
x=[tex]\pm \sqrt{2}[/tex]

Check:
-[tex]\sqrt{2} \ne \sqrt{2}[/tex]

[tex]\sqrt{2}[/tex]=[tex]\sqrt{2}[/tex]

x=tex]\sqrt{2}[/tex]; the other solution is extraneous

[Eigenvalue]

b) y=[tex]\sqrt{4-x²}[/tex]
x=-y
y=-x
-x=[tex]\sqrt{4-x²}[/tex]
Square both sides
x²=4-x²
x²=2
x=[tex]\pm \sqrt{2}[/tex]

Check:
-[tex]\sqrt{2} \ne \sqrt{2}[/tex]

[tex]\sqrt{2}[/tex]=[tex]\sqrt{2}[/tex]

x=tex]\sqrt{2}[/tex]; the other solution is extraneous

[Eigenvalue]

b) y=[tex]\sqrt{4-x²}[/tex]
x=[tex]\frac{1}{x¹⁰}[/tex]
y=[tex]\sqrt{4-(1/x²⁰}[/tex]
This can be simplified, but the answer remains the same

[nycmath]

a) y=[tex]\sqrt{4-x²}[/tex]
y=x
x=[tex]\sqrt{4-x²}[/tex]
square both sides
x²=4-x²
2x²=4
x²=2
x=[tex]\pm \sqrt{2}[/tex]

Check:
-[tex]\sqrt{2} \ne \sqrt{2}[/tex]

[tex]\sqrt{2}[/tex]=[tex]\sqrt{2}[/tex]

x=tex]\sqrt{2}[/tex]; the other solution is extraneous

Nice work as always.

[nycmath]

b) y=[tex]\sqrt{4-x²}[/tex]
x=-y
y=-x
-x=[tex]\sqrt{4-x²}[/tex]
Square both sides
x²=4-x²
x²=2
x=[tex]\pm \sqrt{2}[/tex]

Check:
-[tex]\sqrt{2} \ne \sqrt{2}[/tex]

[tex]\sqrt{2}[/tex]=[tex]\sqrt{2}[/tex]

x=tex]\sqrt{2}[/tex]; the other solution is extraneous

Isn't math great?

[nycmath]

b) y=[tex]\sqrt{4-x²}[/tex]
x=[tex]\frac{1}{x¹⁰}[/tex]
y=[tex]\sqrt{4-(1/x²⁰}[/tex]
This can be simplified, but the answer remains the same

Sometimes the LaTex does not reveal fully. I think this is a site issue. Thank you for your time.